uv'+u'v
x.3sec^2(3x)+tan(3x)
= 3xsec^2(3x)+tan(3x)
Correct.Tan3x + 3xsec^2(3x)
You left a 3 off at the start of your answer.<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{\partial&space;}{\partial&space;x}[xtan3x]&space;=&space;x.3sec^{2}3x+1.tan3x&space;=&space;xsec^{2}3x+tan3x" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{\partial&space;}{\partial&space;x}[xtan3x]&space;=&space;x.3sec^{2}3x+1.tan3x&space;=&space;xsec^{2}3x+tan3x" title="\frac{\partial }{\partial x}[xtan3x] = x.3sec^{2}3x+1.tan3x = xsec^{2}3x+tan3x" /></a>
I don't think this is in 2U?<a href="http://www.codecogs.com/eqnedit.php?latex=\textup{Find&space;the&space;derivative&space;of&space;}&space;5^{x}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\textup{Find&space;the&space;derivative&space;of&space;}&space;5^{x}" title="\textup{Find the derivative of } 5^{x}" /></a>
<a href="http://www.codecogs.com/eqnedit.php?latex=Area&space;=&space;\frac{\pi&space;r^{2}}{4}&space;+&space;\frac{1}{2}.b.h&space;+&space;l.h" target="_blank"><img src="http://latex.codecogs.com/gif.latex?Area&space;=&space;\frac{\pi&space;r^{2}}{4}&space;+&space;\frac{1}{2}.b.h&space;+&space;l.h" title="Area = \frac{\pi r^{2}}{4} + \frac{1}{2}.b.h + l.h" /></a>
It is.. hint: you have to use log laws to change to a base of eI don't think this is in 2U?
Hint: To find the area you integrate the function (from my understanding) so when you're integrating that function, you should end up with the area that function contains.<a href="http://www.codecogs.com/eqnedit.php?latex=Area&space;=&space;\frac{\pi&space;r^{2}}{4}&space;+&space;\frac{1}{2}.b.h&space;+&space;l.h" target="_blank"><img src="http://latex.codecogs.com/gif.latex?Area&space;=&space;\frac{\pi&space;r^{2}}{4}&space;+&space;\frac{1}{2}.b.h&space;+&space;l.h" title="Area = \frac{\pi r^{2}}{4} + \frac{1}{2}.b.h + l.h" /></a>
<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{\pi&space;1^{2}}{4}&space;+&space;\frac{1}{2}.1.1&space;+&space;3.1&space;=&space;\frac{\pi&space;+2&space;+&space;12}{4}&space;=&space;\frac{\pi&space;+&space;14}{4}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{\pi&space;1^{2}}{4}&space;+&space;\frac{1}{2}.1.1&space;+&space;3.1&space;=&space;\frac{\pi&space;+2&space;+&space;12}{4}&space;=&space;\frac{\pi&space;+&space;14}{4}" title="= \frac{\pi 1^{2}}{4} + \frac{1}{2}.1.1 + 3.1 = \frac{\pi +2 + 12}{4} = \frac{\pi + 14}{4}" /></a>
question 2 im stuck
nope thats not itderivate of 5^x = 1/xlogx
is the function pi +14/4 though, the thing we just proved?Hint: To find the area you integrate the function (from my understanding) so when you're integrating that function, you should end up with the area that function contains.
I don't think sois the function pi +14/4 though, the thing we just proved?
im completely stuck -.-I don't think so
is it in the 10th hour? i'm so bad at visualising.A ship anchored in a port has a ladder which hangs over the side. The length of the ladder is 200cm, the distance between each rung in 20cm and the bottom rung touches the water. The tide rises at a rate of 10cm an hour. When will the water reach the fifth rung?
Yeah, that's pretty much the answerYou found the positive area under the curve. When you integrate, anything below the x-axis becomes negative
you must leave spaces between '<' for it to showis it in the 10th hour? i'm so bad at visualising.
Can someone help me with this question:
Find the exact area enclosed by the curve y=sinx and the line y=1/2 for 0<x<2pi
<a href="http://www.codecogs.com/eqnedit.php?latex=\textup{Find&space;the&space;derivative&space;of&space;}&space;5^{x}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\textup{Find&space;the&space;derivative&space;of&space;}&space;5^{x}" title="\textup{Find the derivative of } 5^{x}" /></a>